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Answer 1 · 2017-04-01T17:07:05

The normality is the number of gram equivalents in 1 L of solution.

Equivalent mass

A workable definition of equivalent mass (or equivalent weight) is:

the mass of one reactant in a balanced equation divided by the number of moles of another (this isn't the official definition).

Thus, equivalent mass depends on the reaction involved.

Gram equivalents

For example, in the reaction

$H_{2} {SO}_{4} + 2NaOH \to {Na}_{2} {SO}_{4} + {2H}_{2} O$ $"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O"$

The equivalent mass of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ is its molar mass divided by 2 (the coefficient of $NaOH$ $"NaOH"$ ).

In the reaction

$H_{2} {SO}_{4} + {Ba(OH)}_{2} \to {BaSO}_{4} + {2H}_{2} O$ $"H"_2"SO"_4 + "Ba(OH)"_2 → "BaSO"_4 + "2H"_2"O"$

The equivalent mass of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ is its molar mass divided by 1 (the coefficient of ${Ba(OH)}_{2}$ $"Ba(OH)"_2$ ).

An equivalent mass of a compound is often called an equivalent (eq) or a gram equivalent (geq).

If we call the number of moles of the other compound the n-factor, we can write

$color(blue)(bar(ul(|color(white)(a/a) "geq" = "molar mass"/"n-factor"color(white)(a/a)|)))" "$

Thus, for titration with $"NaOH"$ ,

$"1 geq of H"_2"SO"_4 = ("98.08 g H"_2"SO"_4)/2 = "49.04 g H"_2"SO"_4$ .

For precipitation $"BaSO"4$ ,

$"1 geq of H"_2"SO"_4 = ("98.08 g H"_2"SO"_4)/1 = "98.08 g H"_2"SO"_4$ .

Normality

Normality is defined as the number of gram equivalents of a compound in 1 L of solution.

$color(blue)(bar(ul(|color(white)(a/a)N = "geq"/"litres"color(white)(a/a)|)))" "$

Thus, a $1 N$ solution of sulfuric acid for titration with $"NaOH"$ will contain 49.04 g of the acid.

A $1 N$ solution of sulfuric acid for precipitation of $"BaSO"_4$ will contain 98.08 g of the acid.